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A slightly more elaborate setup which showcases the weirdness of the Axiom of Choice. Consider a group of countably infinite wizards w_1, w_2,... A king has p
by malaya_zemlya 4y ago
A slightly more elaborate setup which showcases the weirdness of the Axiom of Choice.
Consider a group of countably infinite wizards w_1, w_2,... A king has put a number 0 or 1 on their backs and ordered the wizards to stand in a single file, such that wizard w_n can only see the numbers on the backs of all wizards with higher numbers (w_n+1, w_n+2...). As is the tradition, the king gave them a challenge: if you can guess the number on your back, your life is spared, otherwise off with your head. The wizards can discuss their actions beforehand, but no communications is allowed after the numbers have been assigned.
The wizards cannot see their own numbers, they cannot exchange information, yet, if we assume the Axiom of Choice, they still can work out a strategy, such that all but a finite number of wizards survive.
The trick is to break the space of all 0-1 sequences into equivalence classes, whre X is equivalent to Y if they disagree only in finite number places. (in other words, they fully agree starting from some index k). That's an equivalence relation, as you can easily verify.
Then the wizards use the Axiom of Choice to remember one representative sequence from each class (there are infinitely many such sequences, but our wizards are infinitely wise). During the test, each wizard would look at what numbers they see in front of them, pick the representative sequence that is equivalent to what they are seeing, and guess the number that corresponds to their place in the sequence.
All wizards will agree on what representative sequence will need to be picked (because the king's sequence belongs to a single equivalence class), and that sequence will agree with the king's sequence everywhere except for a finite number of places. Boom!
- voldacar 4y ago>Then the wizards use the Axiom of Choice to remember one representative sequence from each class I don't think AoC is enough here becuase the wizards need to evaluate the choice function to get the representative from each equivalence class. AoC merely states that there is some function that maps each equivalence class to some element within that equivalence class. It doesn't tell you how to actually evaluate it, which is what the wizards need to do.
- CaptainNegative 4y agoWhat's the role of restricting the number domain to binary here? Can't they each have a real-valued number on their back (or worse, something from a larger set) with the same argument still going through?
- IngoBlechschmid 4y agoIndeed, it works as long as the class of labels forms a set instead of a proper class.